![]() It's difficult to say what the designer intended.I am noob,still learning the basics,right now I am trying to understand how Darlington transistor pair works.Its probably simplest thing in electrical engineering after two resistor voltage divider yet I cant understand it.Specificaly I am looking at live graphical animation in the preset of falstad free online circuit builder and simulator. It guarantees that Q1 is well into deliberate conduction to cause Q2 to turn on. It is probably to make sure Q2 doesn't come on due to leakage of Q1. R1 (1K) what major role does it play in first circuit ? A buck switcher controlled to regulate current will be significantly smaller, cheaper, and require much less heat management. 45 W might be doable if you get the right transistor with a large heat sink and forced air cooling. If the 300 mA value is to be believed, that's still 45 W. That could be nearly 150 W if the current is really supposed to be regulated to 1 A. No matter what transistor or other pass element you use, a linear regulator dissipates the voltage it drops times the current thru it. However, the real problem is that the high power dissipation is still there. Note that with 2.3 Ω sense resistor against a 700 mV reference, the current will be regulated to 300 mA, not 1 A. That works as long as you are OK with the various inaccuracies and temperature dependencies. You are using the B-E junction of a PNP transistor as the voltage reference to compare the voltage across a current sense resistor against. ![]() Your new circuit has at least somewhat predictable current regulation. The controller of the buck switcher uses that as feedback to regulate the current. You need a small current sense resistor somewhere, but that doesn't need to drop more than a few 100 mV. You haven't told us what this is really for, but I would look into a buck switcher run to control the current directly. Even if it could, it would take a lot of space and cost a lot. That's well beyond what a single transistor with good heat sink can dissipate. With 150 V in, the pass element will drop 149 V. Another really big problem is the power dissipation. It will be very difficult to set the current to any one value with any kind of accuracy. Then you really don't want to use this topology. Input Voltage from Car battery varies from 30V -150V. The purpose of this circuitry is to limit current through load resistor for 1 A when 150 Volts appears on the car battery. The gains also vary as a function of temperature, collector current, and other parameters it's hard to know the effect of. Those can vary considerably between devices, even from the same manufacturing lot. Q3 makes a reasonably good voltage-controlled current sink, but the final load current limit depends too much on the gains of Q1 and Q2. If your goal, for example, is to make a current source, then this is not a good circuit. The above describes how the circuit works, but it's not clear what you are actually trying to accomplish. Hand waving of undescribed dimensionless quantities is not appreciated, and may result in downvoting and/or closing of questions with such issues in the future. It's not clear what that is, and what "150" is supposed to mean. This was all assuming reasonable positive voltage from V3. Note that if the darlington doesn't saturate, then there can substantial voltage across it, which causes significant power dissipation, which might destroy Q2. If the load doesn't draw this amount, then the darlington transistor saturates, as would any BJT in its place. So roughly at first approximation, the maximum load current this circuit can provide is V2 minus the B-E drop of Q3, divided by R11, times the gain of Q1, times the gain of Q2. In this case the gain is roughly the gains of Q1 and Q2 multiplied together. Just like a regular PNP, the maximum collector current is the base current times the gain. Q1 and Q2 act like a single high-gain PNP transistor with a high saturation voltage. Note that this is fairly independent of the voltage on the collector of Q3. To first approximation, Q3 sinks V2 minus a junction drop, divided by R11. Due to the transistor's gain, most of that comes from the collector, not the base. That current obviously flows out of the emitter of Q3. That voltage on R11 causes a particular current to flow. Whatever voltage is applied to its base appears across R11 minus the B-E drop.
0 Comments
Leave a Reply. |
Details
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |